Proving a known function of position via Chain Rule

AI Thread Summary
The discussion revolves around using the Chain Rule to prove that velocity can be expressed as a function of position when acceleration is known. The integral equation provided relates the change in velocity to the integral of acceleration over position. Participants express confusion regarding the correct application of the Chain Rule and the integral setup, with one noting a mistake in the equation's sign. A hint is given to consider the left-hand side as an integral of velocity with respect to velocity. The conversation highlights the importance of correctly applying mathematical principles to derive the desired relationship in rectilinear motion.
kylera
Messages
40
Reaction score
0

Homework Statement


Use the Chain Rule to prove that for rectilinear motion, when the acceleration is a known function of position, you can find the velocity as a function of position via the integral

\frac{v^{2}-v_{0}^{2}}{2} = \int^{s}_{s_{0}}a(s)ds


Homework Equations


v^{2}=v_{0}^{2}\times2as


The Attempt at a Solution


I took the left fraction, substituted v^2, simplified and got as. I let A be as, then took dA to get a da. Now I'm stuck.
 
Physics news on Phys.org
I think you simplified wrong. It's a multipication sign not addition. See what I mean?
 
Hold up, I wrote the question on the board wrong -- it is supposed to be a plus for the relevant equation part.
 
I'm not exactly sure what they're asking here. For constant accelerations, you "relevant equation" is basically the answer, assuming you swap out the "x" for a "+" and take a square root. What's throwing me is the request for proof by chain rule.
 
Chain Rule

kylera said:

Homework Statement


Use the Chain Rule to prove that for rectilinear motion, when the acceleration is a known function of position, you can find the velocity as a function of position via the integral

\frac{v^{2}-v_{0}^{2}}{2} = \int^{s}_{s_{0}}a(s)ds


Homework Equations


v^{2}=v_{0}^{2}\times2as


The Attempt at a Solution


I took the left fraction, substituted v^2, simplified and got as. I let A be as, then took dA to get a da. Now I'm stuck.

Hi kylera! :smile:

You were asked to use the Chain Rule. So …

Hint: the LHS is ∫vdv. So use the Chain Rule on dv. :smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top